Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

Show Tags

05 Nov 2011, 13:23

41

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

68% (01:36) correct
32% (01:50) wrong based on 1166 sessions

HideShow timer Statistics

Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete because of A's help?

A) (x – y)/(x + y) B) x/(y – x) C) (x + y)/(xy) D) y/(x – y) E) y/(x + y)

Re: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

Show Tags

05 Nov 2011, 14:34

1

This post received KUDOS

3

This post was BOOKMARKED

Not very sure but just thought of the following way:-

A's rate is 1/x B's rate is 1/y.

when they work together, then work is done in time xy/(x+y). In this time, work done by b is x/(x+y), by a is y/(x+y) and total work is 1. So the work which b did not do that is the work done by a should be represented as the fraction of total work a's work/total work = [y/x+y] / 1 = y/(x+y)

Re: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

Show Tags

05 Nov 2011, 19:02

3

This post received KUDOS

Plugging in numbers, Let A rate = 3 B rate = 2 To finish a work of 10, B would work 5 hours to finish it, but only 4 hours with help of A. So the fraction left is 2/5

Plug in numbers y/(x+y) = 2/5

Hence E

Hope that helps
_________________

Aim for the sky! (800 in this case) If you like my post, please give me Kudos

Re: Machine A can complete a certain job in x hours. Machine B can complet [#permalink]

Show Tags

05 Nov 2011, 19:09

2

This post received KUDOS

1

This post was BOOKMARKED

Another approach is algebraic.

Machine A completes a job in x hours so the rate is 1/x Machine B complets a job in y hours so the rate is 1/y Collectively 1/x + 1/y = x + y /(xy) for both working together

To get the fraction of what B doesn't have to complete because of A's help is simply to get the fraction of A's workrate when they are both working together

Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

27 Jan 2012, 16:46

2

This post received KUDOS

19

This post was BOOKMARKED

Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete because of A's help?

A. (x – y)/(x + y) B. x/(y–x) C. (x+y)/xy D. y/(x-y) E. y/(x+y)

Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete because of A's help?

A. (x – y)/(x + y) B. x/(y–x) C. (x+y)/xy D. y/(x-y) E. y/(x+y)

Working together A and B complete the job in \(\frac{xy}{x+y}\) hours (as time is a reciprocal of rate then take the reciprocal of combined rate, which is \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\));

Now, in \(\frac{xy}{x+y}\) hours A will complete \(\frac{1}{x}*\frac{xy}{x+y}=\frac{y}{x+y}\) part of the job (rate*time=job) and this will be the part which B will not have to complete.

Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

24 Jun 2012, 19:55

Hi Bunuel,

Can you help me understand that when we want to calculate the work just done by A and not B of the total work which in this case we assume it as 1. Then why do we multiply the rate of only A with the total time?

I can't get my head around this concept. Is it one of those things that I just accept and move on.

Take an example to understand it. A takes 2 hrs (= x) to complete a work alone and B takes 2 hrs (= y) to complete a work alone. Together, they will complete the work in 1 hr = [xy/(x+y) = 2*2/(2+2)] So A does 1/2 of the work in 1 hr (A's rate of work) and B does 1/2 of the work in 1 hr (B's rate of work).

So when they were working together, each needed to work for only 1 hr. How much work was done by A in that one hr? 1*(1/2) = (1/2) So because of A's help, B doesn't need to do half of the work i.e. work done by A.
_________________

Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

19 Sep 2012, 17:21

2

This post received KUDOS

My solution might be very simple:

If we pick numbers like x = 3hrs and y = 4hrs then it takes 7hrs to complete two units for both of them. (We can use two units because we are looking for fractional work, not a total quantity #). If it takes 7hrs between the two of them then Ma does 3/7 or the work and Mb does 4/7 of the work. This means that Mb is not doing 3/7 of the work and you can plug in your answer choices to find out which one gives you x=3, y=4 -> 3/7

Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

15 Nov 2012, 01:45

VeritasPrepKarishma wrote:

Take an example to understand it. A takes 2 hrs (= x) to complete a work alone and B takes 2 hrs (= y) to complete a work alone. Together, they will complete the work in 1 hr = [xy/(x+y) = 2*2/(2+2)] So A does 1/2 of the work in 1 hr (A's rate of work) and B does 1/2 of the work in 1 hr (B's rate of work).

So when they were working together, each needed to work for only 1 hr. How much work was done by A in that one hr? 1*(1/2) = (1/2) So because of A's help, B doesn't need to do half of the work i.e. work done by A.

This explanation is vague since you picked 1/2. We cannot see how it works because it does not matter what you do with x and y it's around 1/2. I see that formula is absolutely correct but don't understand why if the value of 1/x is 2/6 and 1/y is 3/6, the portion which y should not do comes out as 3/5 not 2/5.

This explanation is vague since you picked 1/2. We cannot see how it works because it does not matter what you do with x and y it's around 1/2. I see that formula is absolutely correct but don't understand why if the value of 1/x is 2/6 and 1/y is 3/6, the portion which y should not do comes out as 3/5 not 2/5.

The explanation was meant for saswani who had trouble understanding "Then why do we multiply the rate of only A with the total time?" It doesn't matter what the numbers are - as long as they are easy to work with - you can make out what's going on.

As for your question "why if the value of 1/x is 2/6 and 1/y is 3/6, the portion which y should not do comes out as 3/5 not 2/5."

If x = 3 and y = 2, A does 1/3rd work in 1 hr and B does 1/2 work in 1 hr. When they work together, they complete the work in 1/(1/2 + 1/3) = 6/5 hrs

In 6/5 hrs, A does (1/3)*(6/5) = 2/5 work and B does (1/2)*(6/5) = 3/5 work

So B does NOT need to do 2/5 of the work. Answer still (E).
_________________

Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

07 Jan 2013, 04:09

Can someone please explain the formula xy / x+y to state the work done together. I thought I had a well understanding of this concept, however when I saw this question, I lost it completely. Thanks in advance

Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

07 Jan 2013, 04:22

1

This post received KUDOS

The amount of work that A can do in 1 hour ( or the rate at which A works ) is 1/x, since A takes x hours to complete the whole work. Similarly the amount of work that B can do in 1 hour is 1/y. Now, total work that they can do, when working together, in one hour is 1/x + 1/y. You can also write it as 1/W = 1/x + 1/y = (x + y ) / xy. 1/W is the rate at which they are working. Hence, to complete the total work they need xy / (x + y) hours. Please let me know if anything is not clear.

Can someone please explain the formula xy / x+y to state the work done together. I thought I had a well understanding of this concept, however when I saw this question, I lost it completely. Thanks in advance

Re: Machine A can complete a certain job in x hours. Machine B [#permalink]

Show Tags

07 Jan 2013, 04:57

Iam really confused right now, I understand everything completely except for this essential part. I went through the stuff you told me Bunuel and I understand everything stated in there. But how do you come from 1/x + 1/y --> (x+y) / xy??? Thanks in advance

Iam really confused right now, I understand everything completely except for this essential part. I went through the stuff you told me Bunuel and I understand everything stated in there. But how do you come from 1/x + 1/y --> (x+y) / xy??? Thanks in advance

You mean you don't get how 1/x + 1/y equals to (x+y)/xy?

The same way as \(\frac{1}{2}+\frac{1}{3}=\frac{3+2}{2*3}=\frac{5}{6}\).
_________________